SOLUTION: Time in the air. A ball is tossed into the air from a height of 12 feet at 16 ft/sec. How long does it take to reach the earth?

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Question 321378: Time in the air. A ball is tossed into the air from a height
of 12 feet at 16 ft/sec. How long does it take to reach the
earth?

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Time in the air. A ball is tossed into the air from a height
of 12 feet at 16 ft/sec. How long does it take to reach the
earth?
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Acceleration due to gravity on Earth is ~-16ft/sec/sec. Negative because it's down.
Height as a function of time is
h(t) = -16t^2 + vt + h0, t in seconds, h in feet
v is the initial velocity, 16 ft/sec and h0 is the initial height, 12 feet
h(t) = -16t^2 + 16t + 12
It impacts when h = 0
-16t^2 + 16t + 12 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case -16x%5E2%2B16x%2B12+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2816%29%5E2-4%2A-16%2A12=1024.

Discriminant d=1024 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-16%2B-sqrt%28+1024+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2816%29%2Bsqrt%28+1024+%29%29%2F2%5C-16+=+-0.5
x%5B2%5D+=+%28-%2816%29-sqrt%28+1024+%29%29%2F2%5C-16+=+1.5

Quadratic expression -16x%5E2%2B16x%2B12 can be factored:
-16x%5E2%2B16x%2B12+=+%28x--0.5%29%2A%28x-1.5%29
Again, the answer is: -0.5, 1.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-16%2Ax%5E2%2B16%2Ax%2B12+%29

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Ignore the -0.5 seconds
t = 1.5 seconds